PAT (Advanced Level)-1107 Social Clusters(並查集)
1107 Social Clusters (30 分)
When register on a social network, you are always asked to specify your hobbies in order to find some potential friends with the same hobbies. A social cluster is a set of people who have some of their hobbies in common. You are supposed to find all the clusters.
Input Specification:
Each input file contains one test case. For each test case, the first line contains a positive integer N (≤1000), the total number of people in a social network. Hence the people are numbered from 1 to N. Then N lines follow, each gives the hobby list of a person in the format:
Ki: hi[1] hi[2] ... hi[Ki]
where Ki (>0) is the number of hobbies, and hi[j] is the index of the j-th hobby, which is an integer in [1, 1000].
Output Specification:
For each case, print in one line the total number of clusters in the network. Then in the second line, print the numbers of people in the clusters in non-increasing order. The numbers must be separated by exactly one space, and there must be no extra space at the end of the line.
Sample Input:
8 3: 2 7 10 1: 4 2: 5 3 1: 4 1: 3 1: 4 4: 6 8 1 5 1: 4
Sample Output:
3 4 3 1
社交圈 是由人組成且圈中的人彼此有愛好相同
模擬樣例:愛好分為1 3 4 5 6 7 8 10 這幾種
一共8個人(從0開始記) 共同愛好4下 有1 3 5 7這四個人 共同愛好3下 有 024 這三個人 雖然6與2 有共同愛好5 但6無法歸入024的圈中 因此 共 3個圈
用並查集 新開一個數組為愛好 初始化為0 若i人愛好t hobby[t]=i 此時這個愛好t 由i做老大 後來有人再喜歡這個愛好自動歸順i人
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
vector<int> clus,f;
int hobby[1001];
int getf(int v)
{
return f[v]==v?v:f[v]=getf(f[v]);
}
void merge(int u,int v)
{
int t1,t2;
t1=getf(u);
t2=getf(v);
if(t1!=t2)
{
f[t1]=t2;
}
}
int main()
{
int n;
scanf("%d",&n);
clus.resize(n+1);
f.resize(n+1);
for(int i=1;i<=1001;i++)f[i]=i;
for(int i=1;i<=n;i++)
{
int k;
scanf("%d:",&k);
for(int j=0;j<k;j++)
{
int t;
scanf("%d",&t);
if(hobby[t]==0)
hobby[t]=i;
merge(i,getf(hobby[t]));
}
}
int cnt=0;
for(int i=1;i<=n;i++)
{
clus[getf(i)]++;
}
for(int i=1;i<=n;i++)
{
if(clus[i])cnt++;
}
sort(clus.begin(),clus.end(),greater<int>());
printf("%d\n",cnt);
for(int i=0;i<cnt;i++)
{
printf("%d",clus[i]);
if(i!=cnt-1)printf(" ");
}
}